Codeforces Educational Round 9 Problem D

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CFEdu9D Longest Subsequence

Solution Sketch

If a number $b$ can be divided by a number $a$, then this number $a$ can be contained in $b$’s LCM subsequence.

Recall how we build up prime table using Sieve of Eratosthenes, we mark all multiples of a certain number $x$ if $x$ is a prime. We can use the same technique for this problem!

AC Code

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#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n, m;
    scanf("%d %d", &n, &m);
    int inp[n];
    map<int, int> cnt;
    for (int i = 0; i < n; i++) {
        int tmp;
        scanf("%d", &tmp);
        inp[i] = tmp;
        if (tmp <= m) // major performace improvement
            cnt[tmp]++;
    }
    int res[m + 1];
    memset(res, 0, sizeof(res));
    for (auto i : cnt) {
        if (i.first > m)
            break;
        if (i.second == 0)
            continue;
        for (int j = i.first; j <= m; j += i.first) {
            res[j] += i.second;
        }
    }
    int mx = 0, num = -1;
    for (int i = 0; i <= m; i++) {
        // lcm many not be in cnt!! a * b < m will be fine!
        if (mx < res[i]) {
            num = i;
            mx = res[i];
        }
    }
    if (num == -1) {
        printf("1 0\n");
    } else {
        printf("%d %d\n", num, mx);
        vector<int> ans;
        for (int i = 0; i < n; i++) {
            if (num % inp[i] == 0) {
                ans.push_back(i + 1);
            }
        }
        sort(ans.begin(), ans.end());
        for (auto i : ans)
            printf("%d ", i);
    }
    return 0;
}